Difference between revisions of "2016 AMC 10B Problems/Problem 1"

(Problem)
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<math>\textbf{(A)}\ 1\qquad\textbf{(B)}\ 2\qquad\textbf{(C)}\ \frac{5}{2}\qquad\textbf{(D)}\ 10\qquad\textbf{(E)}\ 20</math>
 
<math>\textbf{(A)}\ 1\qquad\textbf{(B)}\ 2\qquad\textbf{(C)}\ \frac{5}{2}\qquad\textbf{(D)}\ 10\qquad\textbf{(E)}\ 20</math>
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==Solution==
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<math>\frac{frac{1}{a}*(2+frac{1}{2})}{a}</math> then becomes <math>frac{frac{5}{2}}{a^{2}}</math> which is equal to <math>frac{5}{2]*2^{2}</math> = <math>{(D)}\ 10\qquad\textbf{(E)}\ 20</math>
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Solution by ngeorge

Revision as of 09:17, 21 February 2016

Problem

What is the value of $\frac{2a^{-1}+\frac{a^{-1}}{2}}{a}$ when $a= \frac{1}{2}$?

$\textbf{(A)}\ 1\qquad\textbf{(B)}\ 2\qquad\textbf{(C)}\ \frac{5}{2}\qquad\textbf{(D)}\ 10\qquad\textbf{(E)}\ 20$



Solution

$\frac{frac{1}{a}*(2+frac{1}{2})}{a}$ then becomes $frac{frac{5}{2}}{a^{2}}$ which is equal to $frac{5}{2]*2^{2}$ (Error compiling LaTeX. ! Missing } inserted.) = ${(D)}\ 10\qquad\textbf{(E)}\ 20$

Solution by ngeorge

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